# MATH 1100: QUANTITATIVE ANALYSIS TEST #3 (VERSION A) ```MATH 1100: QUANTITATIVE ANALYSIS
TEST #3 (VERSION A)
1. Suppose that the demand for a product is given by pq + p = 4.
(a) Find the elasticity when p = \$2.
(b) What type of elasticity is this (unitary, elastic, inelastic)?
(c) How would revenue be affected by an increase in price?
Solution:
(a) Solving for q in terms of p:
q=
4−p
.
p
To find the elasticity
p dq
q dp
4
=
.
4−p
η=−
At p = 2 this gives η = 2.
(b) The demand is elastic (η &gt; 1).
(c) When the demand is elastic, revenue decreases with price increase.
2. Total revenue is R(x) = 18x − x2 where x is the number of units sold.
(a) What number of units sold maximizes revenue?
(b) What is the maximum revenue?
Solution:
(a) x = 9 (right in between the two zeros of R(x), and you don’t even
need calculus to see that).
(b)
R(x) = R(9) = 81.
3. (a) Find critical points, values at the critical points, local maxima, local
minima and inflection points of
3
C(x) = x3 − x2 − 18x + 6
2
defined for −4 ≤ x ≤ 4.
(b) Graph C(x) for −4 ≤ x ≤ 4.
Solution:
(a)
C 0 (x) = 3x2 − 3x − 18
= 3(x + 2)(x − 3)
Date: October 18, 2001.
1
2
MATH 1100: QUANTITATIVE ANALYSIS
TEST #3 (VERSION A)
20
10
–4
–3
–2
–1
1
x
2
3
4
0
–10
–20
–30
so the critical points are x = −4, −2, 3, 4. The critical values are
C(−4) = −10
C(−2) = 28
69
1
C(3) = −
= −34 −
2
2
C(4) = −26.
So x = −4 is a local min, x = −2 is a global max, x = 3 a global min
and x = 4 is a local max.
1
00
C (x) = 6x − 3 = 6 x −
.
2
So the inflection is at x = 1/2.
(b) See figure 3b.
4. Suppose that the number x of units produced per month depends on the
number y of thousands of dollars invested with
x = 3y + 2y 2 .
At what rate will production increase if \$10000 is invested and the the
investment capital is increasing at a rate of \$10 per month?
Solution:
dx
dx dy
=
dt
dy dt
dy
= (3 + 4y)
dt
MATH 1100: QUANTITATIVE ANALYSIS
TEST #3 (VERSION A)
3
Plug in y = 10 (thousand dollars) and dy/dt = 1/1000 (thousands of dollars
per month):
dx
1
= (3 + 4(10))
dt
1000
43
=
.
1000
5. Suppose
x + exy = y 2 .
Find
where x = 0 and y = 1.
Solution:
dy
dy
1 + exy y + x
= 2y .
dx
dx
Plug in x = 0 and y = 1,
dy
= 1.
dx
dy
dx
```